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Velocity versus time form acceleration versus time

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data

    I have some data from a lab conducted with a accelerometer to collect the acceleration and I am trying to figure out how to create a v vs t graph from the a vs t graph. When I create the a vs t graph in excel its very crazy looking because of the change in acceleration at every 0.1 s from 0 s to 79.6 s. I know the v vs t is constructed from the area under the a vs t graph. The speed of the object is 0 km/h to 40 km/h = 0 m/s to 11.1111 m/s. I have every point of the instantaneous acceleration at every 0.1 s.

    my a vs t graph:

    my acceleration data looks like this from 0 s to 79.6 s:

    2. Relevant equations

    [tex]\Delta v = \int a dt[/tex]

    3. The attempt at a solution

    well I know the time is 0 s to 79.6 s and the velocity is 0 m/s to 11.1111 m/s so I think I can integrate?

    [tex]\Delta v = \int a dt[/tex]
  2. jcsd
  3. Nov 1, 2009 #2


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    Integration is actually a discrete sum in this case. You don't have an analytical function that you can integrate. Just multiply the acceleration at a given point in time by the time interval Δt = 0.1 s and call that Δv. Add all the Δv's from time zero to the time of interest to get the velocity at the time of interest.
  4. Nov 2, 2009 #3
    I used this formula:

    [tex]v_{n} = v_{n-1} + a_{n}(0.1)[/tex]

    which is what you said but would the position be:

    [tex]x_{n} = x_{n-1} + v_{n-1}(0.1) + (0.5)a_{n}(0.1)^{2}[/tex]

    [tex]y_{n} = y_{n-1} + v_{n-1}(0.1) + (0.5)a_{n}(0.1)^{2}[/tex]

    for every point?
  5. Nov 2, 2009 #4


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    It will (probably) be better to use the "other kinematic equation"

    \Delta x_{n} = \frac{v_{n}^2-v_{n-1}^2}{2a_{n}}
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